Chapter 10.7, Problem 26E

To determine

To draw:

The graph of each equation and the value of $\theta$ to the nearest degree.

Answer to Problem 26E

  $\theta ={22.5}^0$ .

Explanation of Solution

Given:

The quadratic equation

  $9x^2+4xy+5y^2-40=0$ .

Concept used:

For the general equation

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ .

The angle rotating $\theta$ about the origin

  $\theta =\frac{\pi }{4}$ If $A=C$

  $\tan 2\theta =\frac{B}{A-C},\text{if}A\ne C$

The number is the square root of the quadratic formula.

  $B^2-4AC$

Calculation:

The quadratic equation

  $9x^2+4xy+5y^2-40=\mathrm{0.......................}\left(1\right)$

Comparing the quadratic equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

  $A=9,B=4,C=5$

The angle rotating $\theta$ about the origin

  $\theta =\frac{\pi }{4}$ If $A=C$

  $\begin{array}{l}\tan 2\theta =\frac{B}{A-C},\text{if}A\ne C\\ \tan 2\theta =\frac{4}{9-5}\\ \tan 2\theta =1\\ 2\theta ={\tan }^{-1}\left(1\right)\\ \theta =\frac{{45}^0}{2}\\ \theta ={22.5}^0\end{array}$

Draw the graph

  $9x^2+4xy+5y^2-40=0$